QUESTION NO 1(A):Let the electric potential at a point in space is zero, does it means that the electric filed intensity is zero at that point? And what happens to electric potential if the electric field intensity is zero at a point in space? Explain both cases. (B): What do you mean by electric flux density? And how you will define (in words) the Gauss's law in term of electric flux de nsity? Under what conditions is Gauss's law useful in calculating the electric field intensity of a charge distribution?
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SUBJECT:ELECTROMAGNETIC FIELD THEORY
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- Gauss’s law also helps us locate charge on conductors. There are several distinct pieces we need forthis, listed below. Explain briefly why is each of them true.(a) The electric field inside a conductor (in the conducting material) is zero if the conductor is instatic equilibrium (charges are not moving or at least not accelerating.)(b) The total charge on an isolated conductor (or, really, any object) is constant.(c) The net charge on the wall of a cavity in a conductor is equal to the opposite of the total chargethat is floating inside the cavity.(d) Any excess charge on a conductor is all at the surfaces - including both the outer surface of theconductor and any inner surfaces of cavities - but never in the material.(e) The charge on the outer surface of a conductor is equal to the total charge on the conductorand inside any cavities in it.This one is tougher! A sphere of radius r has charge q. (a) What is the infinitesimal increase in clectric potential energy dU if an infinitesimal amount of charge dq is brought to infinity to the surface of the sphere? (b) An uncharged sphere can acquire a total charge Q by the transfer of charge dq over and over and over. Use your answer to part a to find an cxpression for the potential energy of a uniformly-charged sphere of radius R with total charge Q. Answer: U = 3_1 Q² 5 4tc0 R' (c) Your answer to part b is the amount of energy nceded to assemble a charged sphere. It is often called the self-energy of the sphere. What is the self-energy of a proton, assuming it to be a charged sphere with a diamcter of 1.0 x 10 15 m?Problem 1: A spherical conductor is known to have a radius and a total charge of 10 cm and 20uC. If points Aand B are 15 cm and 5 cm from the center of the conductor, respectively. If a test charge, q = 25mC, is to bemoved from A to B, determine the following:b.) The electric potential energy at B
- (a) What is capacitance? You are given a parallel-plate capacitor with vacuum between them, now if you replace the vacuum (K=1) by other dielectric material with K value is more than one, then how will the capacitance change? explain on the basis of polarization. (b) In Fig. 1, two parallel-plate capacitors (with air between the plates) are connected to a battery. Capacitor 1 has a plate area of 1.5 cm? and an electric field (between its plates) of magnitude 2000 V/m. Capacitor 2 has a plate area of 0.70 cm? and an electric field of magnitude 1500 V/m. What is the total C2 Figure 1 charge on the two capacitors?Final answers should be in two decimal places only. Include proper substitution and cancellation of units. An electron is placed in a uniform electric field of (given E) directed north created by two horizontal parallel plates that are (given distance) apart. The electron accelerates from rest, starting at the negative plate. What is the net force and acceleration acting on the electron? How long did it take the electron to get from one side of the plate to the other? Given: Electri field (nN/C): 30 Distance (fm): 133Imagine that two parallel plates are charged to achieve a voltage difference Vo between the plates. They have area A and spacing d. The plates are then electrically isolated so that no charge can be added or taken away. A good conductor of thickness d/2 is placed in between the two plates, as shown. Given that the charge on the plates cannot change, what is the electric field in the regions between the plates, but above and below the conductor? (If you are stuck, Gauss’s Law may help.)
- Answer All.Compute for the work done, in millijoules, in moving a 9-nC charge radially away from the center from a distance of 3 m to a distance of 7 m against the electric field inside a solid insulating sphere of radius 11 m and total charge 7 mC.Ans: -8.5199Determine the total potential energy, in microjoules, stored in a parallelepiped of dimensions are 9 m by 6 m by 8 m if the electric field inside is given as E = 17 ar + 19 aθ + 15 aϕ V/m. Use the permittivity of free space as 8.854 × 10-12 F/m.Ans: 1.6734If the electric field in the region is given as E = -cos(θ) sin( 4 Φ) aθ + b cos( 4 Φ) aφ V/m. Determine the potential at point A(4 m, 0.46 rad, 2.07 m), in volts, if the potential at point B(4 m, 1.00 rad, 0.10 m) is 60 volts. The value of b is also the coefficient of Φ.58.4552 Compute for the potential difference, in volts, in moving a charge from A(3, 2, -2) m to B(7, -6, 6) m against the electric field due to a disk charge of radius 9 m on the plane x = 0. The disk has a…Imagine that two parallel plates are charged to achieve a voltage difference Vo between the plates. They have area A and spacing d. The plates are then electrically isolated so that no charge can be added or taken away. A good conductor of thickness d/2 is placed in between the two plates, as shown. How great is the charge density on the surface of the conductor relative to that on either of the plates? Why? (You may use infinite plane approximations for the surfaces.)A conductor that carries a net charge Q has a hollow, empty cavity in its interior. Does the polential vary from point to point within the material of the conductor? What about within the caivity? How does the potential inside the cavity compare to the potential within the material of the conductor? For the toolbar, press ALT+F10 (PC) or AL FN+F10 (Mac). BIUS Paragraph vAV I S.. Arial 14px OWORDS POWERED BY TINY
- Explain with the help of a diagram (if needed) the relationship between the electronic polarizability of molecules and the ability of a high value dielectric material (εr = 10) to either increase or decrease the charge storing capacity of a parallel-plate capacitor if such a material is inserted within the gap between the two plates. For your answer assume that the typically used dielectric material between the two plates of a capacitor is air (εr = 1).Please explain Keatly 'aind cleardy with Equationo oShowrng the corest process Yand answees I have tilud tuwo way La(6). and only gethe same answeifor 10 Quest ioi (6). Thauk yau!/ An electron with a speed of 3.70 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.74 x 103 N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is 4.17 mm long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region? (a) Number i 0:0142239849 Units meteRs (b) Number i 7.69X 10 - 9 seconds Units (c) Number i -1 i UnitsExplain this statement: “In a static situation, the electric field at the surface of a conductor can have no component parallel to the surface because this would violate the condition that the charges on the surface are at rest.” Would this statement be valid for the electric field at the surface of an insulator? Explain your answer and the reason for any differences between the cases of a conducto and an insulator.